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प्रश्न
Prove that (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B.
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उत्तर
L.H.S. = (1 – cos2A) . sec2B + tan2B (1 – sin2A)
= `sin^2A * 1/(cos^2B) + (sin^2B)/(cos^2B) (1 - sin^2A)` ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - cos^2A = sin^2A)]`
= `(sin^2A)/(cos^2B) + (sin^2B)/(cos^2B) - (sin^2A sin^2B)/(cos^2B)`
= `(sin^2A)/(cos^2B) - (sin^2A sin^2B)/(cos^2B) + (sin^2B)/(cos^2B)`
= `(sin^2A)/(cos^2B) (1 - sin^2B) + tan^2B`
= `(sin^2A)/(cos^2B) (cos^2B) + tan^2B`
= sin2A + tan2B
= R.H.S.
∴ (1 – cos2A) . sec2B + tan2B (1 – sin2A) = sin2A + tan2B
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
